Suppose we wanted to evaluate the double integral $S = \iint_D \cos(x) + y \, dx \, dy$ by first applying a change of variables from $D$ to $R$ : $\begin{aligned} x &= X_1(u, v) = u + \cos(v) \\ \\ y &= X_2(u, v) = u + \sin(v) \end{aligned}$ What is $S$ under the change of variables? Assume $0 < v < \dfrac{\pi}{2}$. If you know an expression within absolute value is non-negative, do not use absolute value at all. $S = \iint_R $ $du \, dv$
If we have a transformation $\bold{X} : R \to D$, then we can rewrite an integral under the change of variables: $ \iint_D f(x, y) \, dA = \iint_R f(\bold{X}(u, v)) | J(\bold{X}) | \, du \, dv$ First we need to find the absolute value of the Jacobian, $|J(\bold{X})|$. $\begin{aligned} |J(\bold{X})| &= \left| \det \begin{pmatrix} \dfrac{\partial X_1}{\partial u} & \dfrac{\partial X_1}{\partial v} \\ \\ \dfrac{\partial X_2}{\partial u} & \dfrac{\partial X_2}{\partial v} \end{pmatrix} \right| \\ \\ &= \left| \det \begin{pmatrix} 1 & -\sin(v) \\ \\ 1 & \cos(v) \end{pmatrix} \right| \\ \\ &= \left| \cos(v) + \sin(v) \right| \end{aligned}$ We are given that $0 < v < \dfrac{\pi}{2}$, and both $\cos(v) > 0$ and $\sin(v) > 0$ in that interval. Therefore, we can simplify $\left| \cos(v) + \sin(v) \right|$ to $\cos(v) + \sin(v)$. Now we substitute $u$ and $v$ in $f(x, y)$. $\begin{aligned} f(x, y) &= f(\bold{X}(u, v)) \\ \\ &= \cos( X_1(u, v) ) + X_2(u, v ) \\ \\ &= \cos(u + \cos(v)) + u + \sin(v) \end{aligned}$ Putting everything together, we get the integral under the change of variables: $ \iint_R (\cos(v) + \sin(v)) \left[ \cos(u + \cos(v)) + u + \sin(v) \right] \, du \, dv$